Răspuns :
Pentru monotonie putem folosi doua lucruri:
Diferenta:
[tex]D=x_{n+1}-x_n[/tex]
Daca e mai mare decat 0, atunci e strict crescator, si daca e mai mic decat 0, atunci e strict descrescator
Raportul:
[tex]R=\frac{x_{n+1}}{x_n}[/tex]
Acelasi lucru si aici, numai ca il compari cu 1, nu cu 0
In cazurile in care nu e specificat domeniul de definitie pentru n, il vom considera mai mare sau egal decat 1
1.
[tex]a_n=\frac{n+1}{2n}\\ D=a_{n+1}-a_n=\frac{n+2}{2n+2}-\frac{n+1}{2n}=\frac{2n(n+2)-(n+1)(2n+2)}{2n(2n+2)}\\\\ D=\frac{2n^2+4n-2n^2-2n-2n-2}{4n(n+1)}=\frac{-2}{4n(n+1)}\\\\ \left\begin{array}{ll} -2\ \ \textless \ \ 0\\ n\geq1\rightarrow 4n(n+1)\ \textgreater \ 0 \end{array}\right]\rightarrow\frac{-2}{4n(n+1)}\ \ \textless \ \ 0\rightarrow \boxed{D\ \ \textless \ \ 0}[/tex]
Sirul este strict descrescator ==> are margine superioara: M = a₁ = 1
n ≥ 1 ==> n + 1 > 0 si 2n > 0 ==> (n+1) / (2n) > 0 (raport de numere pozitive) ==> sirul are margine inferioara: m = 0
0 < an ≤ 1 ==> Sirul este marginit
2.
[tex]a_n=\frac{2n+1}{n+2}\\ D=a_{n+1}-a_n=\frac{2n+3}{n+3}-\frac{2n+1}{n+2}=\frac{(2n+3)(n+2)-(n+3)(2n+1)}{(n+2)(n+3)}\\\\ D=\frac{2n^2+7n+6-2n^2-7n-3}{(n+2)(n+3)}=\frac{3}{(n+2)(n+3)}\\\\ \left\begin{array}{ll} 3\ \textgreater \ 0\\ n\geq1\rightarrow (n+2)(n+3)\ \textgreater \ 0 \end{array}\right]\frac{3}{(n+2)(n+3)}\ \textgreater \ 0\rightarrow \boxed{D\ \textgreater \ 0}[/tex]
Sirul este strict crescator. ==> Are margine inferioara: m = a₁ = 1
[tex]a_n=\frac{2n+1}{n+2}=\frac{n+2+n-1}{n+2}=\frac{n+2}{n+2}+\frac{n-1}{n+2}=1+\frac{n-1}{n+2}[/tex]
n ≥ 1 ==> n - 1 ≥ 0 si n + 2 > 0 ==> Daca n - 1 < n + 2 (ceea ce e adevarat), arunci fractia este subunitara: (n - 1)/(n + 2) < 1 ==> 1 + (n - 1)/(n - 2) < 2 ==> Sirul are margine superioara: M = 2
1 ≤ an < 2 ==> Sirul este marginit
3.
[tex]a_n=\sum^n_{k=1}\frac{1}{k^2}\\ D = a_{n+1}-a_n=\sum^{n+1}_{k=1}\frac{1}{k^2}-\sum^n_{k=1}\frac{1}{k^2}\\\\ D=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}+\frac{1}{(n+1)^2}-(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2})=\frac{1}{(n+1)^2}\\\\ \left\begin{array}{ll}1\ \textgreater \ 0\\ n\geq1\rightarrow (n+1)^2\ \textgreater \ 0\end{array}\right]\rightarrow \frac{1}{(n+1)^2}\ \textgreater \ 0\rightarrow \boxed{D\ \textgreater \ 0}[/tex]
Sirul este strict crescator ==> Are margine inferioara: m = a₁ = 1
Nu stiu sa demonstrez marginirea acestui sir in mod direct, asa ca ma voi folosi de un alt sir a carui marginire o putem calcula:
[tex]x_n=\sum^n_{k=2}\frac{1}{k(k-1)}\\ \text{Vom folosi urmatoarea formula: }\frac{1}{(k-1)k}=\frac{1}{k-1}-\frac{1}{k}\\ \text{Desfasuram suma:}\\ x_n=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{(n-1)n}\\ x_n=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{n-1}-\frac{1}{n}\\\\ \text{Se observa ca se reduc toti termenii, mai putin primul si ultimul}\\ \boxed{x_n=1-\frac{1}{n}\ \textless \ 1}[/tex]
Acum comparam cele 2 siruri:
[tex]x_n=\sum^n_{k=2}\frac{1}{k(k-1)}=\sum^n_{k=2}\frac{1}{k^2-k}\\ a_n=\sum^n_{k=1}\frac{1}{k^2}=1+\sum^n_{k=2}\frac{1}{k^2}\\\\ k\ \textgreater \ 0\rightarrow k^2\ \textgreater \ k^2-k\rightarrow \frac{1}{k^2}\ \textless \ \frac{1}{k^2-k}\rightarrow \sum^n_{k=2}\frac{1}{k^2}+1\ \textless \ \sum^n_{k=2}\frac{1}{k^2-k}+1\\ a_n\ \textless \ x_n+1\ \textless \ 2\rightarrow \boxed{a_n\ \textless \ 2}[/tex]
Sirul are margine superioara: M = 2
1 ≤ an < 2 ==> Sirul este marginit
4.
[tex]a_n=\sum^n_{k=1}\frac{1}{5^k+1}\\ D=a_{n+1}-a_n=\sum^{n+1}_{k=1}\frac{1}{5^k+1}-\sum^n_{k=1}\frac{1}{5^k+1}\\ D=\frac{1}{5^{n+1}+1}+\sum^n_{k=1}\frac{1}{5^k+1}-\sum^n_{k=1}\frac{1}{5^k+1}\\ D=\frac{1}{5^{n+1}+1}\\\\ \left\begin{array}{ll} 1\ \textgreater \ 0\\ n\geq1\rightarrow5^{n+1}\ \textgreater \ 5\rightarrow5^{n+1}+1\ \textgreater \ 0 \end{array}\right]\rightarrow \frac{1}{5^{n+1}-1}\ \textgreater \ 0\rightarrow \boxed{D\ \textgreater \ 0} [/tex]
Sirul este strict crescator ==> Are margine inferioara: m = a₁ = 1/6
Si aici vom face la fel. Ne vom folosi de urmatorul sir:
[tex]x_n=\sum^n_{k=1}\frac{1}{5^k}\\ \text{Folosim formula sumei termenilor unei proresii geometrice}:\\ S=b_1\frac{1-q^n}{1-q}\\\\ x_n=\frac{1}{5^1}+\frac{1}{5^2}+...+\frac{1}{5^n}\\\\ x_n=\frac{1}{5}\frac{1-\frac{1}{5^n}}{1-\frac{1}{5}}=4(\frac{5^n-1}{5^{n-2}})=100-\frac{4}{5^{n-2}}\\ \frac{4}{5^{n-2}}\ \textgreater \ 0\rightarrow100-\frac{4}{5^{n-2}}\ \textless \ 100\rightarrow \boxed{x_n\ \textless \ 100}[/tex]
Comparam cele 2 siruri:
[tex]5^{k}+1\ \textgreater \ 5^k\rightarrow \frac{1}{5^{k}+1}\ \textless \ \frac{1}{5^k}\rightarrow a_n\ \textless \ x_n\ \textless \ 100\rightarrow \boxed{a_n\ \textless \ 100}[/tex]
Sirul are margine superioara: M = 100
1/6 ≤ an < 100 ==> Sirul este marginit
Diferenta:
[tex]D=x_{n+1}-x_n[/tex]
Daca e mai mare decat 0, atunci e strict crescator, si daca e mai mic decat 0, atunci e strict descrescator
Raportul:
[tex]R=\frac{x_{n+1}}{x_n}[/tex]
Acelasi lucru si aici, numai ca il compari cu 1, nu cu 0
In cazurile in care nu e specificat domeniul de definitie pentru n, il vom considera mai mare sau egal decat 1
1.
[tex]a_n=\frac{n+1}{2n}\\ D=a_{n+1}-a_n=\frac{n+2}{2n+2}-\frac{n+1}{2n}=\frac{2n(n+2)-(n+1)(2n+2)}{2n(2n+2)}\\\\ D=\frac{2n^2+4n-2n^2-2n-2n-2}{4n(n+1)}=\frac{-2}{4n(n+1)}\\\\ \left\begin{array}{ll} -2\ \ \textless \ \ 0\\ n\geq1\rightarrow 4n(n+1)\ \textgreater \ 0 \end{array}\right]\rightarrow\frac{-2}{4n(n+1)}\ \ \textless \ \ 0\rightarrow \boxed{D\ \ \textless \ \ 0}[/tex]
Sirul este strict descrescator ==> are margine superioara: M = a₁ = 1
n ≥ 1 ==> n + 1 > 0 si 2n > 0 ==> (n+1) / (2n) > 0 (raport de numere pozitive) ==> sirul are margine inferioara: m = 0
0 < an ≤ 1 ==> Sirul este marginit
2.
[tex]a_n=\frac{2n+1}{n+2}\\ D=a_{n+1}-a_n=\frac{2n+3}{n+3}-\frac{2n+1}{n+2}=\frac{(2n+3)(n+2)-(n+3)(2n+1)}{(n+2)(n+3)}\\\\ D=\frac{2n^2+7n+6-2n^2-7n-3}{(n+2)(n+3)}=\frac{3}{(n+2)(n+3)}\\\\ \left\begin{array}{ll} 3\ \textgreater \ 0\\ n\geq1\rightarrow (n+2)(n+3)\ \textgreater \ 0 \end{array}\right]\frac{3}{(n+2)(n+3)}\ \textgreater \ 0\rightarrow \boxed{D\ \textgreater \ 0}[/tex]
Sirul este strict crescator. ==> Are margine inferioara: m = a₁ = 1
[tex]a_n=\frac{2n+1}{n+2}=\frac{n+2+n-1}{n+2}=\frac{n+2}{n+2}+\frac{n-1}{n+2}=1+\frac{n-1}{n+2}[/tex]
n ≥ 1 ==> n - 1 ≥ 0 si n + 2 > 0 ==> Daca n - 1 < n + 2 (ceea ce e adevarat), arunci fractia este subunitara: (n - 1)/(n + 2) < 1 ==> 1 + (n - 1)/(n - 2) < 2 ==> Sirul are margine superioara: M = 2
1 ≤ an < 2 ==> Sirul este marginit
3.
[tex]a_n=\sum^n_{k=1}\frac{1}{k^2}\\ D = a_{n+1}-a_n=\sum^{n+1}_{k=1}\frac{1}{k^2}-\sum^n_{k=1}\frac{1}{k^2}\\\\ D=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}+\frac{1}{(n+1)^2}-(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2})=\frac{1}{(n+1)^2}\\\\ \left\begin{array}{ll}1\ \textgreater \ 0\\ n\geq1\rightarrow (n+1)^2\ \textgreater \ 0\end{array}\right]\rightarrow \frac{1}{(n+1)^2}\ \textgreater \ 0\rightarrow \boxed{D\ \textgreater \ 0}[/tex]
Sirul este strict crescator ==> Are margine inferioara: m = a₁ = 1
Nu stiu sa demonstrez marginirea acestui sir in mod direct, asa ca ma voi folosi de un alt sir a carui marginire o putem calcula:
[tex]x_n=\sum^n_{k=2}\frac{1}{k(k-1)}\\ \text{Vom folosi urmatoarea formula: }\frac{1}{(k-1)k}=\frac{1}{k-1}-\frac{1}{k}\\ \text{Desfasuram suma:}\\ x_n=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{(n-1)n}\\ x_n=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{n-1}-\frac{1}{n}\\\\ \text{Se observa ca se reduc toti termenii, mai putin primul si ultimul}\\ \boxed{x_n=1-\frac{1}{n}\ \textless \ 1}[/tex]
Acum comparam cele 2 siruri:
[tex]x_n=\sum^n_{k=2}\frac{1}{k(k-1)}=\sum^n_{k=2}\frac{1}{k^2-k}\\ a_n=\sum^n_{k=1}\frac{1}{k^2}=1+\sum^n_{k=2}\frac{1}{k^2}\\\\ k\ \textgreater \ 0\rightarrow k^2\ \textgreater \ k^2-k\rightarrow \frac{1}{k^2}\ \textless \ \frac{1}{k^2-k}\rightarrow \sum^n_{k=2}\frac{1}{k^2}+1\ \textless \ \sum^n_{k=2}\frac{1}{k^2-k}+1\\ a_n\ \textless \ x_n+1\ \textless \ 2\rightarrow \boxed{a_n\ \textless \ 2}[/tex]
Sirul are margine superioara: M = 2
1 ≤ an < 2 ==> Sirul este marginit
4.
[tex]a_n=\sum^n_{k=1}\frac{1}{5^k+1}\\ D=a_{n+1}-a_n=\sum^{n+1}_{k=1}\frac{1}{5^k+1}-\sum^n_{k=1}\frac{1}{5^k+1}\\ D=\frac{1}{5^{n+1}+1}+\sum^n_{k=1}\frac{1}{5^k+1}-\sum^n_{k=1}\frac{1}{5^k+1}\\ D=\frac{1}{5^{n+1}+1}\\\\ \left\begin{array}{ll} 1\ \textgreater \ 0\\ n\geq1\rightarrow5^{n+1}\ \textgreater \ 5\rightarrow5^{n+1}+1\ \textgreater \ 0 \end{array}\right]\rightarrow \frac{1}{5^{n+1}-1}\ \textgreater \ 0\rightarrow \boxed{D\ \textgreater \ 0} [/tex]
Sirul este strict crescator ==> Are margine inferioara: m = a₁ = 1/6
Si aici vom face la fel. Ne vom folosi de urmatorul sir:
[tex]x_n=\sum^n_{k=1}\frac{1}{5^k}\\ \text{Folosim formula sumei termenilor unei proresii geometrice}:\\ S=b_1\frac{1-q^n}{1-q}\\\\ x_n=\frac{1}{5^1}+\frac{1}{5^2}+...+\frac{1}{5^n}\\\\ x_n=\frac{1}{5}\frac{1-\frac{1}{5^n}}{1-\frac{1}{5}}=4(\frac{5^n-1}{5^{n-2}})=100-\frac{4}{5^{n-2}}\\ \frac{4}{5^{n-2}}\ \textgreater \ 0\rightarrow100-\frac{4}{5^{n-2}}\ \textless \ 100\rightarrow \boxed{x_n\ \textless \ 100}[/tex]
Comparam cele 2 siruri:
[tex]5^{k}+1\ \textgreater \ 5^k\rightarrow \frac{1}{5^{k}+1}\ \textless \ \frac{1}{5^k}\rightarrow a_n\ \textless \ x_n\ \textless \ 100\rightarrow \boxed{a_n\ \textless \ 100}[/tex]
Sirul are margine superioara: M = 100
1/6 ≤ an < 100 ==> Sirul este marginit
Vă mulțumim că ați accesat site-ul nostru dedicat Matematică. Sperăm că informațiile furnizate v-au fost utile. Dacă aveți întrebări sau aveți nevoie de asistență suplimentară, nu ezitați să ne contactați. Vă așteptăm cu drag să reveniți și nu uitați să ne salvați la favorite!